Find $\int \dfrac{1}{\sqrt{-x^2-6x+40}}\,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{14}\text{arctan}\left(\dfrac{x+3}{7}\right)+C$ (Choice B) B $\dfrac{1}{14}\text{arcsin}\left(\dfrac{x+3}{7}\right)+C$ (Choice C) C $\text{arcsin}\left(\dfrac{x+3}{7}\right)+C$ (Choice D) D $\text{arctan}\left(\dfrac{x+3}{7}\right)+C$
Explanation: The integrand is in the form $\dfrac{1}{\sqrt{p(x)}}$ where $p(x)$ is a quadratic expression. This suggests that we should rewrite $p(x)$ by completing the square. Specifically, we will rewrite $p(x)$ as $ k^2-(x+ h)^2$. Then, we will be able to integrate using our knowledge of the derivative of the inverse sine function: $\int \dfrac{1}{\sqrt{ k^2-x^2}}\,dx=\text{arcsin}\left(\dfrac{x}{ k}\right)+C$ [Why is this formula true?] By setting $u=x+ h$ and using $u$ -substitution, we get the following formula: $\int \dfrac{1}{\sqrt{ k^2-(x+ h)^2}}\,dx=\text{arcsin}\left(\dfrac{x+ h}{ k}\right)+C$ We start by rewriting $p(x)$ as $ k^2-(x+ h)^2$ : $\begin{aligned} -x^2-6x+40&=40-(x^2+6x) \\\\ &=40+9-(x^2+6x+9) \\\\ &=49-(x+3)^2 \\\\ &={7}^2-(x+{3})^2 \end{aligned}$ Now we can find the integral: $\begin{aligned} &\phantom{=}\int \dfrac{1}{\sqrt{-x^2-6x+40}}\,dx \\\\ &=\int\dfrac{1}{\sqrt{{7}^2-(x+{3})^2}}\,dx \\\\ &=\text{arcsin}\left(\dfrac{x+{3}}{{7}}\right)+C \end{aligned}$ In conclusion, $\int \dfrac{1}{\sqrt{-x^2-6x+40}}\,dx=\text{arcsin}\left(\dfrac{x+3}{7}\right)+C$